\begin{answer}
    Let $l^{(i)} = (o^{(i)} - y^{(i)})^2$. Then

    $$
    \frac{\partial l}{\partial w^{[1]}_{1, 2}} = \frac{1}{m}\Sigma_{i=1}^m\frac{\partial l^{(i)}}{\partial w^{[1]}_{1, 2}}
    $$

    Note that $w^{[1]}_{1, 2}$ is only affecting $h_2^{(i)}$. And thus
    $$
    \frac{\partial l^{(i)}}{\partial w^{[1]}_{1,2}} = \frac{\partial l^{(i)}}{\partial o^{(i)}}\frac{\partial o^{(i)}}{\partial h_2^{(i)}}\frac{\partial h_2^{(i)}}{\partial w^{[1]}_{1, 2}}
    $$

    $$
\frac{\partial l^{(i)}}{\partial w^{[1]}_{1,2}} = \frac{\partial l^{(i)}}{\partial o^{(i)}}\frac{\partial o^{(i)}}{\partial h_2^{(i)}}\frac{\partial h_2^{(i)}}{\partial w^{[1]}_{1, 2}}
$$

Since $l^{(i)} = (o^{(i)} - y^{(i)})^2$,
$$
\frac{\partial l^{(i)}}{\partial o^{(i)}} = 2(o^{(i)} - y^{(i)})
$$
Since $o^{(i)} = \sigma(w^{[2]}_0 + w^{[2]}_1 h_1^{(i)} + w_2^{[2]}h_2^{(i)} + w_3^{[3]}h_3^{(i)})$ ,
$$
\frac{\partial o^{(i)}}{\partial h_2^{(i)}} = o^{(i)}(1 - o^{(i)})w_2^{[2]}
$$
And since $h_2^{(i)} = \sigma(w^{[1]}_{0, 2}  + w^{[1]}_{1, 2}x^{(i)}_1+ w^{[1]}_{2, 2}x^{(i)}_2)$,
$$
\frac{\partial h_2^{(i)}}{\partial w^{[1]}_{1, 2}} = h_2^{(i)}(1- h_2^{(i)}) x_1^{(i)}
$$
Combining them we get
$$
\frac{\partial l}{\partial w^{[1]}_{1, 2}} = \sum_{i=1}^m 2(o^{(i)} - y^{(i)})o^{(i)}(1 -  o^{(i)})\sigma(w^{[1]}_{0, 2}  + w^{[1]}_{1, 2}x^{(i)}_1+ w^{[1]}_{2, 2}x^{(i)}_2)(1 - \sigma(w^{[1]}_{0, 2}  + w^{[1]}_{1, 2}x^{(i)}_1+ w^{[1]}_{2, 2}x^{(i)}_2))w_2^{[2]}x_1^{(i)}
$$
        
 \end{answer}
